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I Rate of Change of Vector in Rotating Frame

  1. Dec 4, 2016 #1
    I recognize the rate of change of a vector in an inertial frame S can be related to the rate of change of the vector in a rotating frame S0 by the equation below taken from my textbook, where Ω is the angular velocity vector. $$\Big(\frac{dQ}{dt}\Big)_{S_{0}}= \Big(\frac{dQ}{dt}\Big)_{S} + \Omega \times Q$$
    However, I am unsure which frame of reference Q refers to in the cross product.
  2. jcsd
  3. Dec 4, 2016 #2
    Define Q.
  4. Dec 4, 2016 #3
    Q is an arbitrary vector such as a point of a ball in space.
  5. Dec 4, 2016 #4
    If Q is a position vector, it's usually measured from a defined origin (in this case probably inertial frame S). Your textbook may or may not have a picture illustrating this.
  6. Dec 4, 2016 #5
    The origins of both frames are coincident and Q can be described in either S or S0, but the textbook doesn't clarify which frame Q is described in for the cross product.
  7. Dec 5, 2016 #6


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    All equations are always in one frame or, even less confusing, you work with invariant objects, namely the vectors themselves, and not with their components. In this case, let's take ##\vec{e}_j## as the time-independent basis vectors in an inertial reference frame and ##\vec{e}_k'## the time-dependent vectors in the rotating frame. Then there's a rotation matrix ##D_{kj}## such that
    $$\vec{e}_k'=D_{kj} \vec{e}_j.$$
    Now take any vector ##\vec{A}##. You have
    $$\frac{\mathrm{d}}{\mathrm{d} t} \vec{A}=\dot{\vec{A}}=\frac{\mathrm{d}}{\mathrm{d} t} (A_k' \vec{e}_k').$$
    Now you have
    $$\dot{\vec{A}}=\dot{A}_k' \vec{e}_k' + A_k' \dot{\vec{e}}_k'. \qquad (1)$$
    But now
    $$\dot{\vec{e}}_k'=\dot{D}_{kj} \vec{e}_j=\dot{D}_{kj} (D^{-1})_{jl} \vec{e}_l'=\epsilon_{klm} \Omega_m' \vec{e}_l', \qquad (2)$$
    where I have used that because of
    $$\hat{D} \hat{D}^T=\hat{1}$$
    the matrix ##\dot{D}_{kj} (D^{-1})_{jl}## is antisymmetric (prove it!). Plugging this in (2) in (1) finally gives
    $$\dot{\vec{A}}=\dot{A}_k' \vec{e}_k' + \epsilon_{klm} A_k' \Omega_m' \vec{e}_l = (\dot{A}_l'+\epsilon_{lmk} \Omega_m' A_k') \vec{e}_l'.$$
    This means that for the components of ##\dot{\vec{A}}## you get
    $$\mathrm{D}_t \underline{A}'=\mathrm{d}_t \underline{A}' + \underline{\Omega}' \times \underline{A}',$$
    where the underlined symbols mean the component-column vectors wrt. the basis ##\vec{e}_k'##.
  8. Dec 5, 2016 #7


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    Choose some fairly easy case, such as a unit vector along the x axis in the ##S_0## frame, and try it both ways.... See which interpretation of the formula gives you a sensible result.

    I'm not suggesting this because it's the best way of resolving the ambiguity in the textbook, but because the exercise will go a long ways towards building your intuition around how these transformations work.
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